Consider Fréchet spaces $\{E_n\}_{n\in\mathbb{N}}$, and let $E = \prod_{n=1}^{\infty} E_n$. Show that $E$ is a Fréchet space.
I'm stuck with proving completeness. I have already verified that $E$ is a metrizable locally convex space, meaning that we can verify completeness using Cauchy sequences (instead of Cauchy filters). Consider a Cauchy sequence $\{x_n\}_{n=1}^{\infty}$ in $E$. Write $x_n=(x_n^{(1)},\dots,x_n^{(\ell)},\dots)$ where each $x_n^{(\ell)}\in E_{\ell}$ and this $\forall n\ge 1$.
Then every $\{x_n^{(\ell)}\}_{n=1}^{\infty}$ is a Cauchy sequence in $E_{\ell}$, and because each $E_{\ell}$ is a Fréchet space (and thus complete, by definition), we have $x_n^{(\ell)}\to x^{(\ell)}\in E_{\ell}$ as $n\to \infty$.
I want to show that $x_n\to (x^{(1)},\dots,x^{(\ell)},\dots)=:x$ as $n\to \infty$. Take a nbh $U$ of $x$ in $E$. Then each projection $p_{\ell}(U)$ of $U$ onto $E_{\ell}$ is a nbh of $x^{(\ell)}$ in $E_{\ell}$. Therefore $\forall \ell\ge 1: \exists N_{\ell}\in\mathbb{N}: \forall n\ge N_{\ell}: x_n^{(\ell)}\in p_{\ell}(U)$.
Can I just say the following: pick $N\in\mathbb{N}$ such that $N\ge N_{\ell},\forall\ell$, then $x_n\in U$ whenever $n\ge N$, proving the convergence in $E$. In a way I'm taking an supremum of a countable infinite set of natural numbers, which seems to be the logical thing to do, but can I do this? (supremum principle?)
Thanks.
Best Answer
Note that the definition of the product topology gives you the following: There are $U_\ell\subseteq E_\ell$ and a finite $I\subseteq\mathbf N$ such that $\prod U_\ell \subseteq U$ and $U_\ell =E_\ell$ for $\ell\not\in I$. Hence you only have to take $\sup_{i\in I} N_i$, leaving you with a finite $N$.